Find $\dfrac{d}{dx}\left[\log\left(\dfrac{1}{1-x}\right)\right]$. Choose 1 answer: Choose 1 answer: (Choice A) A $\dfrac{1-x}{\ln(10)}$ (Choice B) B $-\dfrac{1}{\ln(10)(1-x)}$ (Choice C) C $\dfrac{1}{\ln(10)(1-x)}$ (Choice D) D $\dfrac{1}{1-x}$
Explanation: $\log\left(\dfrac{1}{1-x}\right)$ is a composition of three functions! Let... $u(x)=\log(x)$ $v(x)=\dfrac{1}{x}$ or $x^{-1}$ $w(x)=1-x$... then $\log\left(\dfrac{1}{1-x}\right)=u\biggl(v\Bigl(w(x)\Bigr)\biggr)$. To find $\dfrac{d}{dx}\left[\log\left(\dfrac{1}{1-x}\right)\right]$, we will need to use the chain rule twice! $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left[u\biggl(v\Bigl(w(x)\Bigr)\biggr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot \dfrac{d}{dx}\left[v\Bigl(w(x)\Bigr)\right] \\\\ &=u'\biggl(v\Bigl(w(x)\Bigr)\biggr)\cdot v'\Bigl(w(x)\Bigr)\cdot w'(x) \end{aligned}$ Let's differentiate $u$, $v$, and $w$ : $u'(x)=\dfrac{1}{\ln(10)\cdot x}$ $v'(x)=-\dfrac{1}{x^2}$ $w'(x)=-1$ Now we can plug the equations for $u$, $v$, $w$, $u'$, $v'$, AND $w'$ into the expression we got: = u ′ ( v ( w ( x ) ) ) ⋅ v ′ ( w ( x ) ) ⋅ w ′ ( x ) = 1 ln ( 10 ) ( 1 1 − x ) ⋅ − 1 ( 1 − x ) 2 ⋅ − 1 = ( 1 − x ) ln ( 10 ) ⋅ 1 ( 1 − x ) 2 = 1 ln ( 10 ) ( 1 − x ) \begin{aligned} &\phantom{=}{u'\biggl(v\Bigl(w(x)\Bigr)\biggr)}{\cdot v'\Bigl(w(x)\Bigr)}\cdot{ w'(x)} \\\\ &={\dfrac{1}{\ln(10)\left(\dfrac{1}{1-x}\right)}}\cdot{-\dfrac{1}{(1-x)^2} }\cdot{-1} \\\\ &=\dfrac{\cancel{(1-x)}}{\ln(10)}\cdot \dfrac{1}{(1-x)^\cancel{2}} \\\\ &=\dfrac{1}{\ln(10)(1-x)} \end{aligned} In conclusion, $\dfrac{d}{dx}\left[\log\left(\dfrac{1}{1-x}\right)\right]=\dfrac{1}{\ln(10)(1-x)}$.